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The most should be nitrate ions.

Moles of 600 ml of agno3 solution, i.e. moles of agno3, agno+ and no3- each.

Moles of 400 ml of NaCl solution, i.e. moles of NaCl, Na+, and Cl-Moles.

agno3+nacl==agcl↓+nacl

Because there are ag+ and no3- moles of Na+ and Cl- in the original mixture, and Cl- and Ag+ react to become precipitates after the reaction, Ag+ and Cl- decrease. So after the reaction, the remaining ions of various ions should be sodium ions, molar nitrate ions, molar silver ions, and chloride ions, so it is 0

Original question. Existing molar 600 ml of AgNO3 solution, add 400 ml of molar NaCl solution, after the reaction, which of the following ions is the most in this solution? (All kinds have to be calculated).

1. no3- 2. ag+

I think so, too.

Before the mixing reaction, there are 6mol ml of silver ions and NO3 ions, sodium ions and chloride ions, 1mol silver ions will combine with 1mol chloride ions to form precipitate, the ion concentration in the solution will decrease, the concentration of silver ions will become, chloride ions will be completely consumed, and the concentration of sodium ions will remain unchanged.

It's sodium! Ag and No3 are and Cl are AgnO3 NaCl=AgnCl nano3, but Agcl is a precipitate, a part of it is consumed, and the sodium ions remain unchanged, so there are many sodium ions.

I think it's silver ions.

Believe in yourself! Presumably it may be that the original question is missing a 0, which should be.

It should be that a part of agcll will precipitate, so it cannot be regarded as a solution

1) H+, Cu2+ are oxidizing.

Because the guessing hydrogen element in H+ can change from +1 valence to 0 valence, and the copper element in Cu2+ can change from Sijing +2 valence to 0 valence.

The valency can be reduced, indicating that the element can be reduced, so it has the oxidation of the Spike Hall.

zn + 2h+ =zn2+ +h2↑

cu2+ +fe = fe2+ +cu

The composition of the residue is Na2CO3, NaCl, 2NaHCO3 = Na2CO3 + H2O + CO2 M

168g 106 62g

m( nahco3)

M(NaHCO3) = 42% mass fraction of NaHCO3 in the 42 g mixture

There is a potassium-sodium alloy that reacts with a sufficient amount of water to obtain hydrogen.

The average relative atomic mass of the alloy is.

k 39na 23

The mass fraction of sodium is [ ]100% =

The composition of the residue is sodium carbonate and sodium chloride.

Sodium bicarbonate is heated and decomposed to produce sodium carbonate, water and carbon dioxide, and the reduced mass is the mass of water and carbon dioxide, and the mass of sodium bicarbonate can be found by the difference method. If the sodium bicarbonate is calculated to be 52g, then the mass fraction is 52%.

Set potassium xg and sodium yg. Then x+y= writes the reaction equation, or x39+y23=2 2*2 according to the conservation of the number of transferred electrons. Solve x,y and you're good to go.

The composition of the residue is Na2CO3 and NaCl, and the mass fraction of NaHCO3 in the mixture is 21%.

The residue is sodium carbonate and sodium chloride, only baking soda is decomposed, what is reduced is the mass of water and carbon dioxide produced by the decomposition of baking soda, and the relative molecular mass of water and carbon dioxide is reduced by a quarter of a mol, so there is originally a half of a mol of sodium bicarbonate, that is, 42g of sodium bicarbonate, so the mass fraction is 42/2,,, The amount of the total substance of the second question potassium and sodium mixture is 1mol, and the total mass is, which is found by the cross method, The essence of the cross method is the difference ratio.

Which classmate is so fast==I posted the question before I read it.。

One serving is dropwise with dilute hydrochloric acid until just completely dissolved.

The reaction is. cu（oh）2+2hcl=cucl2+2h2oy 2ycuso4 = cuo + so2+cu（oh）2=cuo+h2o

The total amount of cuo is x+y

It can also be calculated based on the conservation of elements that the total amount of XcuSO4·yCu(OH)2 copper ions is x+y

2y：x+y =3︰2

The calculation shows x=1 y=3

They are -3 +5 0 +4 +2

b Because there is only one chemical tamper in r, +3 in a, r in a, r in c, +1, and r in d, +2

Because Al only has a neutral valence of +3 (OH) and Cl are both -1 valence, so the total anions of this chemical are -1*(n+6-n)=-6, so the cations are +6 in total, and Al is +3, so the remaining Fe is also +3

h is almost always +1, and o is usually -2.

In addition, NH4+, NO3 - you talk about the branch less mark code sensitive symbol.

r+3。Attendant b. OH-SO4 2-NO3-CO3 2-Al+3, Cl-1, OH-1, apparently Fe+3